Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $z = \dfrac{r^2 + r - 90}{-r + 7} \times \dfrac{-5r + 35}{r - 9} $
Explanation: First factor the quadratic. $z = \dfrac{(r - 9)(r + 10)}{-r + 7} \times \dfrac{-5r + 35}{r - 9} $ Then factor out any other terms. $z = \dfrac{(r - 9)(r + 10)}{-(r - 7)} \times \dfrac{-5(r - 7)}{r - 9} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ (r - 9)(r + 10) \times -5(r - 7) } { -(r - 7) \times (r - 9) } $ $z = \dfrac{ -5(r - 9)(r + 10)(r - 7)}{ -(r - 7)(r - 9)} $ Notice that $(r - 7)$ and $(r - 9)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac{ -5\cancel{(r - 9)}(r + 10)(r - 7)}{ -(r - 7)\cancel{(r - 9)}} $ We are dividing by $r - 9$ , so $r - 9 \neq 0$ Therefore, $r \neq 9$ $z = \dfrac{ -5\cancel{(r - 9)}(r + 10)\cancel{(r - 7)}}{ -\cancel{(r - 7)}\cancel{(r - 9)}} $ We are dividing by $r - 7$ , so $r - 7 \neq 0$ Therefore, $r \neq 7$ $z = \dfrac{-5(r + 10)}{-1} $ $z = 5(r + 10) ; \space r \neq 9 ; \space r \neq 7 $